How to find/construct functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$
$begingroup$
Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.
If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?
I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.
real-analysis number-theory
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
$endgroup$
add a comment |
$begingroup$
Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.
If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?
I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.
real-analysis number-theory
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
$endgroup$
1
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36
add a comment |
$begingroup$
Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.
If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?
I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.
real-analysis number-theory
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
$endgroup$
Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.
If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$
After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?
I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.
real-analysis number-theory
real-analysis number-theory
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
edited Dec 12 '18 at 4:26
Pinteco
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
asked Dec 12 '18 at 4:04
PintecoPinteco
731313
731313
1
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36
add a comment |
1
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36
1
1
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.
consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .
Let's define $f$ as follows.
$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$
So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.
And then we define $g$ to be the floor of $e cdot f(n)$ .
$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$
Here are the first few values of $f(n)$
n f(n) exp(f(n))
1 1 2.7182817
2 2 5.4365635
3 3 8.154845
4 7 19.027973
5 39 106.012985
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036215%2fhow-to-find-construct-functions-fn-and-gn-such-that-lim-n-to-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.
consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .
Let's define $f$ as follows.
$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$
So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.
And then we define $g$ to be the floor of $e cdot f(n)$ .
$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$
Here are the first few values of $f(n)$
n f(n) exp(f(n))
1 1 2.7182817
2 2 5.4365635
3 3 8.154845
4 7 19.027973
5 39 106.012985
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
$endgroup$
add a comment |
$begingroup$
Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.
consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .
Let's define $f$ as follows.
$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$
So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.
And then we define $g$ to be the floor of $e cdot f(n)$ .
$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$
Here are the first few values of $f(n)$
n f(n) exp(f(n))
1 1 2.7182817
2 2 5.4365635
3 3 8.154845
4 7 19.027973
5 39 106.012985
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
$endgroup$
add a comment |
$begingroup$
Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.
consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .
Let's define $f$ as follows.
$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$
So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.
And then we define $g$ to be the floor of $e cdot f(n)$ .
$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$
Here are the first few values of $f(n)$
n f(n) exp(f(n))
1 1 2.7182817
2 2 5.4365635
3 3 8.154845
4 7 19.027973
5 39 106.012985
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
$endgroup$
Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.
consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .
Let's define $f$ as follows.
$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$
So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.
And then we define $g$ to be the floor of $e cdot f(n)$ .
$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$
Here are the first few values of $f(n)$
n f(n) exp(f(n))
1 1 2.7182817
2 2 5.4365635
3 3 8.154845
4 7 19.027973
5 39 106.012985
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
edited Dec 12 '18 at 5:29
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
answered Dec 12 '18 at 5:21
Gregory NisbetGregory Nisbet
573312
573312
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036215%2fhow-to-find-construct-functions-fn-and-gn-such-that-lim-n-to-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16
$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17
$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24
$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36